Key Features of Graphs

A graph tells the story of a function at a glance: where it crosses the axes, where it peaks, where it levels off. In IB Mathematics you’ll often get the graph from your GDC (graphic display calculator), and the skill being tested is reading off the key features correctly and copying them onto paper so that someone else can see them. This page shows you which features to look for, how to find each one with technology, and how to turn a screen into a clear, labelled sketch.

Key ideas

Draw versus sketch

IB questions use two different command words, and they mean different things.

Sketch: show the general shape and the key features, with the axes labelled. It doesn’t have to be to scale, but the features have to be in the right places relative to each other.
Draw: make an accurate graph to scale, usually on axes you’re given, with points plotted correctly and joined by a smooth curve (or ruled straight lines).

In both cases, label the axes and label every key feature, usually with its coordinates.

The key features

Feature	What it is	How to find it on a GDC
$y$ -intercept	where the graph crosses the $y$ -axis: $(0, f(0))$	evaluate $f(0)$ , or trace to $x = 0$
zeros ( $x$ -intercepts)	values of $x$ with $f(x) = 0$ ; these are the roots of the equation $f(x) = 0$	the “zero” or “root” tool
local maximum / minimum	the top of a “hill” or bottom of a “valley”; the vertex of a parabola is one of these	the “maximum” / “minimum” tool
symmetry	a line of symmetry like $x = h$ (parabolas), symmetry in the $y$ -axis, or rotational symmetry about a point	look at the graph; check a pair of points like $f(h - 1)$ and $f(h + 1)$
vertical asymptote	a line $x = a$ the graph shoots up or down beside	where the formula is undefined; check the table near $x = a$
horizontal asymptote	a line $y = c$ the graph levels off toward as $x$ gets very large or very negative	the table for large $\lvert x \rvert$ , e.g. $x = 1000$
period	for a repeating graph, the horizontal length of one cycle	distance between two neighbouring maximums

A maximum value or minimum value is a $y$ -value. The point where it happens is a coordinate pair. “The maximum value is $3.08$ ” and “the maximum point is $(-0.155, 3.08)$ ” are both correct answers to different questions, so read carefully.

Transferring a graph from screen to paper

When you copy a graph from your GDC:

Use the same window. Note the $x$ - and $y$ -ranges on the screen (or the domain the question gives) and mark a scale on each axis.
Find and label the key points. Use the GDC tools to get exact or 3 s.f. coordinates of the intercepts, maximums and minimums, and the endpoints if the domain is restricted.
Draw asymptotes as dashed lines and write their equations.
Copy the shape. Make the curve pass through your labelled points and have the right behaviour at the ends: approaching an asymptote, stopping at an endpoint, or heading off the screen.

Intersections with technology

To find where $y = f(x)$ and $y = g(x)$ meet, graph both and use the GDC’s “intersect” tool. The $x$ -coordinates of the intersection points are the solutions of the equation $f(x) = g(x)$ . (Another way: graph $y = f(x) - g(x)$ and find its zeros.) This works for equations you can’t solve by hand, such as $5 - x^2 = e^x$ .

Sums and differences of functions

The graph of $y = f(x) + g(x)$ is made by adding the $y$ -values of the two graphs at each $x$ . On a GDC, enter $f$ and $g$ as two functions, then enter a third as their sum (for example Y3 = Y1 + Y2) and graph it. Some things to notice:

where $g(x) = 0$ , the sum has the same value as $f(x)$
where one function is very large (near an asymptote), the sum is too, so the sum keeps that asymptote
the zeros of $f(x) - g(x)$ are exactly the $x$ -coordinates where $f$ and $g$ intersect

There’s more on combining functions on the page about adding and subtracting functions.

Worked examples

Example 1: Transferring a cubic from a GDC

Sketch the graph of $f(x) = x^3 - 3x^2 - x + 3$ for $-2 \le x \le 4$ , labelling all intercepts, the local maximum and minimum, and the endpoints.

Solution. Graph $f$ on your GDC with window $-2 \le x \le 4$ and $-16 \le y \le 16$ (the endpoints tell you how tall the window needs to be).

Endpoints: $f(-2) = -8 - 12 + 2 + 3 = -15$ and $f(4) = 64 - 48 - 4 + 3 = 15$ .
$y$ -intercept: $f(0) = 3$ .
Zeros (GDC “zero” tool): $x = -1$ , $1$ and $3$ . Check by factoring: $f(x) = (x + 1)(x - 1)(x - 3)$ .
Local maximum (GDC “maximum” tool): $(-0.155, 3.08)$ to 3 s.f.
Local minimum (GDC “minimum” tool): $(2.15, -3.08)$ to 3 s.f.

A sketch copied from a GDC: the axes have scales, and every key point is labelled.

Notice that the endpoints are drawn as solid dots, because the domain includes $x = -2$ and $x = 4$ . With a restricted domain the graph stops there; it doesn’t carry on.

Example 2: Asymptotes and symmetry

Let $g(x) = \dfrac{x^2}{x^2 - 4}$ . Use technology to find the key features of the graph, and describe its symmetry.

Solution.

Vertical asymptotes: the denominator is $0$ when $x^2 = 4$ , so at $x = -2$ and $x = 2$ . The GDC table confirms it: $g(1.99) \approx -99.3$ and $g(2.01) \approx 101$ .
Horizontal asymptote: for large $\lvert x \rvert$ , $g(1000) \approx 1.000004$ , so $y = 1$ .
Intercepts: $g(0) = 0$ , so the graph passes through the origin, and that’s its only zero.
Turning point: the GDC “maximum” tool on the middle section gives a local maximum at $(0, 0)$ .
Symmetry: $g(-x) = \dfrac{(-x)^2}{(-x)^2 - 4} = g(x)$ , so the graph is symmetric in the $y$ -axis (the function is even).

The graph has three pieces: a middle “upside-down bowl” between the asymptotes with its top at the origin, and two outer branches that come down from high up beside $x = \pm 2$ and level off just above $y = 1$ .

Range: $\{y \in \mathbb{R} \mid y \le 0 \text{ or } y \gt 1\}$ . (The middle piece has $y \le 0$ ; the outer branches have $y \gt 1$ .)

Example 3: Intersection in context

Town A has a population of $1200$ and grows by $150$ people a year, so $A(t) = 1200 + 150t$ . Town B has a population of $800$ and grows by $12\%$ a year, so $B(t) = 800(1.12)^t$ . Here $t$ is the time in years. When will the two towns have the same population?

Solution. We need $A(t) = B(t)$ , an equation you can’t rearrange by hand. Graph $y = 1200 + 150x$ and $y = 800(1.12)^x$ on your GDC with a window like $0 \le x \le 20$ , $0 \le y \le 6000$ , and use the “intersect” tool:

t = 11.4 \text{ years (3 s.f.)}, \qquad \text{population} \approx 2910

(The GDC also finds an intersection at a negative $t$ , about $-4.96$ , but that’s before the model starts, so we reject it.)

Check: $A(11.4) = 1200 + 150(11.4) = 2910$ and $B(11.4) = 800(1.12)^{11.4} \approx 2912$ . These agree to 3 s.f.; the small difference comes from rounding $t$ . ✓

Example 4: Graphing a sum of functions

Let $f(x) = x^2$ and $g(x) = \dfrac{2}{x}$ . Graph $h(x) = f(x) + g(x)$ and state its key features.

Solution. Enter Y1 $= x^2$ , Y2 $= 2/x$ and Y3 = Y1 + Y2, and graph all three.

Vertical asymptote: $x = 0$ . $g$ has one there, and adding $x^2$ (which is close to $0$ near $x = 0$ ) doesn’t cancel it.
No horizontal asymptote: for large $\lvert x \rvert$ , $\dfrac{2}{x}$ is tiny and $h(x)$ behaves like $x^2$ , so the graph keeps rising.
Local minimum (GDC): $(1, 3)$ . Check: $h(1) = 1 + 2 = 3$ .
Zero (GDC): $x = -1.26$ (3 s.f.). Exactly, $x^2 + \dfrac{2}{x} = 0$ gives $x^3 = -2$ , so $x = -\sqrt[3]{2}$ .
No $y$ -intercept, because $h(0)$ is undefined.

h(x) = x^2 + \dfrac{2}{x}

(blue) is built by adding the heights of the orange and green graphs.

Common mistakes

Copying the shape but not the labels. A sketch without labelled intercepts, turning points and asymptotes loses most of the marks. Write the coordinates next to each key point, and the equation next to each asymptote.

Drawing past a restricted domain. If the question says $-2 \le x \le 4$ , the graph starts and stops there. Mark the endpoints and label them.

Mixing up the maximum value and the maximum point. The maximum value is the $y$ -coordinate only. If the question asks for the point, give both coordinates.

Trusting the default window. The standard window can hide features: a turning point off the top of the screen, or two zeros so close together they look like one. Zoom in or change the window until you’re sure you’ve seen everything, and use the table to check what happens for large $\lvert x \rvert$ .

Drawing an asymptote as part of the curve. On some calculators, a steep line appears where the graph jumps across a vertical asymptote. It isn’t part of the graph. Draw the asymptote as a dashed line and keep the curve off it.

Keeping intersection points that don’t fit the context. A GDC finds every intersection in the window, including negative times or sizes. Check each one against the domain of the model.

Practice

1. (Warm-up) Let $f(x) = (x - 2)^2 - 9$ . State the vertex, the equation of the axis of symmetry, the zeros and the $y$ -intercept.

Solution

Vertex $(2, -9)$ ; axis of symmetry $x = 2$ .

Zeros: $(x - 2)^2 = 9$ , so $x - 2 = \pm 3$ , giving $x = -1$ and $x = 5$ .

$y$ -intercept: $f(0) = 4 - 9 = -5$ .

2. (Warm-up) State the equations of the asymptotes of $y = 3 + \dfrac{2}{x + 1}$ , and find both intercepts.

Solution

Vertical asymptote $x = -1$ (the fraction is undefined there). Horizontal asymptote $y = 3$ (the fraction gets close to $0$ for large $\lvert x \rvert$ ).

$y$ -intercept: $y = 3 + \dfrac{2}{1} = 5$ .

$x$ -intercept: $3 + \dfrac{2}{x + 1} = 0$ gives $\dfrac{2}{x + 1} = -3$ , so $x + 1 = -\dfrac{2}{3}$ and $x = -\dfrac{5}{3}$ .

3. (Warm-up) The height of a point on a turning wheel is $h(x) = 2\sin(30x) + 3$ metres, where $x$ is in seconds and the angle is in degrees, for $0 \le x \le 24$ . State the period, the maximum and minimum values, and the times when the maximum happens.

Solution

Period: $\dfrac{360}{30} = 12$ seconds.

Since $\sin$ goes between $-1$ and $1$ : maximum value $2(1) + 3 = 5$ m, minimum value $2(-1) + 3 = 1$ m.

The maximum happens when $30x = 90$ or $450$ , so at $x = 3$ s and $x = 15$ s (one period apart).

4. (Core) Use technology to find, to 3 s.f., the zeros and the coordinates of all local maximums and minimums of $f(x) = x^4 - 4x^2 + x + 1$ .

Solution

Graph $f$ with a window like $-3 \le x \le 3$ , $-6 \le y \le 6$ .

Zeros: $x = -2.06$ , $-0.396$ , $0.694$ and $1.76$ .

Local minimums: $(-1.47, -4.44)$ and $(1.35, -1.62)$ . Local maximum: $(0.126, 1.06)$ .

Notice the graph is not symmetric, even though it looks a bit like a “W”: the $+x$ term tilts it, so the two minimums have different heights.

5. (Core) Find the coordinates of the points of intersection of $y = 5 - x^2$ and $y = e^x$ , to 3 s.f.

Solution

Graph both and use the “intersect” tool twice:

(-2.21,\ 0.110) \quad\text{and}\quad (1.24,\ 3.46)

Check the second point: $5 - 1.241^2 \approx 3.46$ and $e^{1.241} \approx 3.46$ . ✓

6. (Core) Sketch a possible graph of a function $f$ with all of these features: domain $\{x \in \mathbb{R} \mid x \ge -4\}$ ; $f(-4) = -2$ ; zeros at $x = -3$ and $x = 4$ ; a local maximum at $(1, 5)$ ; $y$ -intercept $4$ ; and $f(x)$ approaches $-1$ as $x$ gets very large.

Solution

One possible sketch: start with a solid dot at $(-4, -2)$ . Rise through the zero $(-3, 0)$ and the $y$ -intercept $(0, 4)$ to the local maximum $(1, 5)$ . Then fall through the zero $(4, 0)$ and keep going down, levelling off just above (or below) the dashed horizontal asymptote $y = -1$ on the right.

Label all five points, the asymptote $y = -1$ , and both axes. Many different curves fit this description; any smooth curve with exactly these features is correct.

7. (Core) A ball is thrown upward from a balcony. Its height in metres after $t$ seconds is $h(t) = -4.9t^2 + 12t + 1.5$ .

(a) Find the maximum height and when it happens.
(b) Find when the ball hits the ground.
(c) State a suitable domain for the model and a GDC window to view it.

Solution

(a) GDC “maximum” tool (or the vertex formula $t = -\dfrac{12}{2(-4.9)}$ ): $t = 1.22$ s, with maximum height $8.85$ m (3 s.f.).

(b) GDC “zero” tool: $t = 2.57$ s (3 s.f.). The other zero, $t \approx -0.119$ , is before the ball was thrown, so reject it.

(c) Domain $\{t \in \mathbb{R} \mid 0 \le t \le 2.57\}$ . A window like $0 \le x \le 3$ and $0 \le y \le 10$ shows the whole flight.

8. (Challenge) Let $f(x) = 2^x$ and $g(x) = x^2$ .

(a) Graph $d(x) = f(x) - g(x)$ and find its zeros to 3 s.f.
(b) How many times do the graphs of $f$ and $g$ intersect? Explain using part (a).
(c) Find the local maximum and minimum points of $d$ .

Solution

(a) Enter Y1 $= 2^x$ , Y2 $= x^2$ , Y3 = Y1 − Y2 and use the “zero” tool on Y3: $x = -0.767$ , $x = 2$ and $x = 4$ . (Exact checks: $2^2 = 4 = 2^2$ and $2^4 = 16 = 4^2$ .)

(b) Three times. The zeros of $f(x) - g(x)$ are exactly the $x$ -values where $f(x) = g(x)$ . You need a wide enough window to see the third one: on a window that stops at $x = 3$ , you’d miss $x = 4$ .

9. (Challenge) Let $f(x) = \dfrac{3x^2 - 2}{x^2 - x - 6}$ .

(a) Find the equations of all asymptotes.
(b) Show that the graph crosses its horizontal asymptote, and find where.

Solution

(a) $x^2 - x - 6 = (x - 3)(x + 2)$ , and the numerator isn’t $0$ at $x = 3$ or $x = -2$ , so the vertical asymptotes are $x = 3$ and $x = -2$ . For large $\lvert x \rvert$ the $x^2$ terms dominate, so $f(x) \approx \dfrac{3x^2}{x^2} = 3$ : the horizontal asymptote is $y = 3$ . (A GDC table agrees: $f(1000) \approx 3.003$ .)

(b) Solve $f(x) = 3$ :

\begin{aligned} 3x^2 - 2 &= 3(x^2 - x - 6) \\ 3x^2 - 2 &= 3x^2 - 3x - 18 \\ 3x &= -16 \\ x &= -\frac{16}{3} \end{aligned}

So the graph crosses $y = 3$ at $\left(-\dfrac{16}{3}, 3\right)$ . A horizontal asymptote only describes the far-left and far-right behaviour; the graph is allowed to cross it elsewhere.